Compendium 4-03:
Problem is to show that if \(g(x)=\int_\pi^{2\pi}\frac{\sin(xt)}tdt,\)then g'(1)=0.
Compendium solution is complete, except you should justify that the last integral is zero. (What is sin at those two values?)
But, note a shortcut: the integral is of the form \(\int_a^b\frac{H(xt)}t dt,\) and differentiating wrt. x we get \(\int_a^bh(xt) dt\) where H'=h. Inserting x=1, we get the integral of H', and the answer is H(b)-H(a). Here we need not know anything about how to (anti)differentiate trigonometric functions.
Compendium 4-04:
Problem is to show that if \(y(t)=\int_0^t\sin^2(t+x)dx,\)then \(\ddot y(t)=6 \sin(2t) \cos(2t)+2t-4y.\)
See compendium solution. Notice that x is the variable of integration, and you shall differentiate wrt. t. After differentiating twice, you should arrive at
\(\ddot y=6 \sin(2t) \cos(2t)+2\int_0^t\left[\cos^2(t+x)-\sin^2(t+x)\right]dx.\) Here you need to use the identity cos2+sin2=1 to eliminate cos2 and get
\(\ddot y=6 \sin(2t) \cos(2t)+2\int_0^t1dx - 2\int_0^t2\sin^2(t+x)dx=6 \sin(2t) \cos(2t)+2t-4y.\)
Compendium 5-05:
(a) Problem: general solution of \(\dot x+2x=2.\)
You could use the Math 2 formula, but you should definitely take note of the "general solution of homogeneous + particular solution" approach. The corresponding homogeneous equation has general solution Ce-2t, and a particular solution is 1.
(b) Problem: find w such that \(\qquad\ddot w+2\dot w = 2,\quad w(0)=0,\qquad w(-\tfrac12)=\tfrac12-e.\)
Put w'=x. Then w'=1+Ce-2t, and integrating from 0 to t we get w(t)=t+K(1-e-2t) with K to fit to w(-1/2). Doing so, we get 1/2-e = -1/2+K(1-e) and K=1.
Compendium 6-08:
General solution of (i) g''=-g'/4 and (ii) g''(t)=-2g'(t)/(t+1).
Both of these are first-order in g', and can be solved for g' using Math 2. Then integrate g' to get g. Notice however that the first equation can be solved as a second-order linear with constant coefficients, yielding the characteristic equation r2+r/4=0 and roots 0 and -1/4.
(Can the second? Not constant coeff's. You can transform it to an Euler differential equation, but that is not curriculum.)
Compendium 6-10:
The differential equation \(\tfrac12\sigma^2x^2V''(x)+\mu xV'(x)-\rho V(x)=w-x\) (note: to be solved for V(x), no "x(t)" here).
This differential equation is not curriculum in itself, i.e., you will not be given this equation and asked "Solve." But once you have solution candidates, it is curriculum to be able to construct the general solution. Here you are not given any candidate for a particular solution, so (b) and thus (c) will be a bit of a stretch despite the hint, but are IMHO OK for a seminar problem.
(a) The solution of the corresponding homogeneous equation: Insert for the suggestion, cancel the x-dependence and get a quadratic for the exponent: \(h(a)=\frac12\sigma^2a(a-1)+\mu a-\rho=0\) . You will get the answer at the end of the compendium. (You can see from h, before solving out, that there must be two distinct real roots: because the constants are positive, h is a convex quadratic in a, and h(0)<0. So, there is one positive and one negative root.)
(b), (c): A reasonable guess would be a first-degree polynomial px+1. Then fit coefficients.
As discussed with you, I include the following for completeness, as the phenomenon is something that must be considered for the exam relevant case with constant coefficients: would we not get division by zero? The answer is that we would, if px or q were solutions of the homogeneous equation; that is, if one of the roots were a=0 or a=1. But h(0)<0 as mentioned in part (a), and \(h(1)=\mu-\rho\), which was stated to be nonzero.
Compendium 6-12 (the part-and-a-half assigned for this week):
\(\dot p(t)=\beta\int_0^t\left[a-c+(d-b)p(\tau)\right]e^{-\alpha(t-\tau)}d\tau\)
(a) Problem: a 2nd order diff.eq. (1) for p.
The compendium solution tells you to use the Leibniz rule. Alternatively, one can use the form \(\dot p(t)=e^{-\alpha t}\int_0^t f(\tau)d\tau\) and one can use the product rule and then insert.
(b) Equilibrium point = constant solution. For a constant solution, all derivatives are zero - that is, delete the first two terms of (1). (By this week's lectures, you should be able to solve the rest as well!).
Compendium 6-02:
This requires this week's lectures. Form \(\ddot x+\tfrac72\dot x-2x=f(t), \)with f=0 in part (a) and t+sin t in part (b).
(a) This is homogeneous. Characteristic equation: \(r^2+\tfrac72r-2\) with roots \(r=-\tfrac74\pm\sqrt{(\tfrac{7}4)^2+2}=-\tfrac74\pm\sqrt{\tfrac{81}{16}}=\frac{-7\pm9}4\), that is, -4 and 1/2. General solution as given in the compendium (since f=0).
(b) We already have the general solution to the corresponding homogeneous equation, so we only need to add a particular solution u*. With the form suggested, we get \(\dot u^*=A+C\cos t-D\sin t\)and \(\ddot u^*=-\big[C\sin t+ D\cos t\big]. \)Inserting, we shall match \(-C\sin t-D\cos t +\tfrac72\left[A+C\cos t-D\sin t\right]-2\left[At+B+C\sin t+D\cos t\right]\)to t + sin t. Constants should zero out, so \(\tfrac72 A-2B=0;\) first-order terms should match 1, so A=-1/2 and thus B=7/8. Then cos coefficients should zero out, so \(-D+\tfrac72C-2D=0;\)while sin coefficients should match 1, so \(-C-\tfrac72D-2C=1.\)Solve for (C,D) and get the solution in the compendium.
NOTE: Part (b) was "kind to you" in that the form was suggested, and in particular: that the "t" part on the RHS needs At+B (with a constant term) and that a "sin" part needs both a cos and a sin. A version "with a trap" could be as follows: (a) solve x''+7x'/2 - 2x = 5; (b) explain how you would go forth to solve if the "5" in the RHS were replaced by "sin t".
Compendium 6-01:
Was not outright assigned. Will be assigned for the next seminar.